Switching on a template literal expression does not narrow the interpolated union variable

まだ誰も着手していません。

### 🔎 Search Terms

template literal, template expression, switch, narrowing, control flow analysis, interpolated variable, union, as const

### 🕗 Version & Regression Information

This is the behavior in every version tried (4.1.5, 5.9.2, 7.0.2; not expressible before 4.1), and I (had claude) review the FAQ for entries about type narrowing and template literal types

### ⏯ Playground Link

https://www.typescriptlang.org/play/?ts=5.9.2#code/CYUwxgNghgTiAEYD2A7AzgF3gMyUgXPAERQAOpEIR8APsaSLNXUQEZQodREDcAUHzQB3AJYYwAC3gAKAAYASAN64kAX3ikRIWfChpEqTAEp4ivvAuI9CEuUoatRfOcuvk6LGQohCt79QBeHDx+V1cAenD4AD0AfngQAA8GMAwQYGCkeAwszhgYJCFsrL9KXl1UgFcoCAgAT3hgJBB9FCQsPIKhFwtWOCgAa35VAUiDFAwCiEJhMUkRFABzeFRMxpE4VPr4TsL9PQTk8DTgQVFxKWkVEzM3a2IvMucwy3dMXTsfB8-AzNDLPqMIZ8EZAA

### 💻 Code

```ts
declare const foo: "apple" | "pear" | "banana";

switch (`${foo} pie` as const) {
case "apple pie":
const apple: "apple" = foo;
// ^? expected foo to narrow to "apple"; actually does not narrow
break;
}

// control: switching on foo directly narrows as expected
switch (foo) {
case "apple":
const apple: "apple" = foo;
break;
}
```

### 🙁 Actual behavior

Inside `case "apple pie":`, `foo` is still typed `"apple" | "pear" | "banana"`, so `const apple: "apple" = foo` fails with:
```
error TS2322: Type '"apple" | "pear" | "banana"' is not assignable to type '"apple"'.
```

This is wrong because the case can only be reached when `foo === "apple"`. The switch subject `${foo} pie as const` is typed as `"apple pie" | "pear pie" | "banana pie"`, and each case label corresponds to exactly one value of foo. The compiler already computes that correspondence, but control flow analysis doesn't propagate the match back to `foo`. The equivalent `if ((${foo} pie as const) === "apple pie")` fails the same way, while switching on foo directly narrows as expected.

### 🙂 Expected behavior

`foo` should narrow to `"apple"` inside case `"apple pie"`, and correspondingly in the other cases.

Since `foo`'s type is a finite union of string literals, `${foo} pie` evaluates to a distinct literal for each member. The mapping from case label back to foo's value is unambiguous and already known to the checker, which is what makes the subject type `"apple pie" | "pear pie" | "banana pie"` in the first place.

Matching a case label should therefore narrow `foo` the same way switch (foo) does, analogous to how narrowing already works backwards through other expression forms like `typeof foo === "string"` and `switch (true)` (TS 5.3)

### Additional information about the issue

I understand this may be a design limitation rather than a bug, since narrowing generally only tracks references and the switch subject here is an expression. But the checker already computes the exact one-to-one mapping from `foo`'s members to the subject's members when it constructs the type `"apple pie" | "pear pie" | "banana pie"`, so the information needed to narrow is present.

Related but distinct issues found while searching: #46045, #53887, #42007, #20949, #51426 (none cover narrowing the interpolated variable of a template literal expression used as a switch/equality subject).

This issue was drafted using the assistance of LLMs (claude), though everything was read, directed, and reviewed by me before submitting this issue through the github website.

リンクされた TypeScript Playground の再現から始め、template-literal switch と等価性の挙動を foo に対する直接の switch と比較してください。checker に既にある union および template-literal narrowing の経路を追跡してください。各 case で foo が一致する literal に narrowing され、直接の switch の制御にリグレッションがない状態になれば完了です。

索引モデルが issue の本文から書いたものです。